__Values__
x= |
Deflection (mm) |
Moment (kNm) |
Shear force (kN) |

__Ultimate Limit State calculations (MRT):__
__Parameters:__
Concrete type:

=> So this means that fck =

N/mm

^{2}
fcd = fck * 0.85 / 1.5 =

N/mm

^{2}.

The steeltype selected for this beam is

N/mm

^{2}.

fyd = fyk / 1.15 =

N/mm

^{2}.

The environmental factor =

, so C

_{,min,dur} =

mm (EN 1992-1-1 Table 4.5N)

__Calculation of the concrete cover thickness:__
First we calculate "e" assuming no shear rebar is present.

e1 = max(C_{,min,dur} or 1,1*) + 10 mm + 1,1*/2 = mm.

Next we check the influence of shear rebar.

e2 = max(C_{,min,dur} or 1,1*) + 10 mm + 1,1* + 1,1*/2 = mm.

Our final e is the max of these 2 values: e = max (e1,e2) = mm.

Once we know the value for e, calculating the effective height "d" is easy:

d = H - e where H is the height of the beam.

Hence: d = mm - mm = mm.

In an effort to calculate the effective height of the beam, the centre of gravity for the rebar group needs to be calculated.

First we calculate "e" assuming no shear rebar is present for the lowest row of rebar.

e1 = max(C_{,min,dur} or 1,1*) + 10 mm + 1,1*/2 = mm.

Next we check the influence of shear rebar.

e2 = max(C_{,min,dur} or 1,1*) + 10 mm + 1,1* + 1,1*/2 = mm.

Our final e for the bottom row of rebar is the max of these 2 values: e = max (e1,e2) = mm.

Next the acceptable distance between the two rows of rebar needs to be calculated.

This is the maximum of the following factors: 20mm, diameter of the bar, max grain size + 3 mm.

A side not that the diameter of the bar is the biggest of the two if the 2 rows have a different diameter rebar.

The distance of the second row is then e + half of the diameter of the rebar of the first row + max distance in between 2 rows + half of the diameter of the second row.

These distances (e for row 1, e+0,5*d+distance in between+0,5*d for the second row) are then multiplied by the respective total surface-area amount of rebar / row.

This is summed and divided by the total amount of surface area of rebar in the beam (on the tension side).

In our case the numerator is mm^{3} and the denominator is mm^{2}.

When divided by each other we find that e_{total} is mm.

Once we know the value for e, calculating the effective height "d" is easy:

d = H - e where H is the height of the beam.

Hence: d = mm - mm = mm.

__Calculation of the main rebar requirements due to bending moment My__
The defining moment on the tension side is

kNm.

μ = My*10

^{6} / (B*d

^{2}*fcd) =

.

ω = 1 - √(1-2*μ) =

.

As

_{vaad} = ω*B*d*fcd/fyd =

mm

^{2}.

Now that we know the required amount of reinforcement, we need to check the minimum amount of reinforcement necessary.

As

_{min} = max(0.26*fctm*B*d/fyk or 0.0013*B*d)

If we calculate fctm by 0,3*fck

^{2/3} =

N/mm

^{2}.

We find that As

_{min} is

mm

^{2}.

As a result the minimum required amount of reinforcement defines the amount of reinforcement to be used.

So technically it is not necessary to define main rebar in this case.

The minimum required amount of reinforcement is not defining for the amount of reinforcement to be used. As

_{vaad} defines the total amount of reinforcement to be used.

We have mm^{2} present, so the amount of main rebar is higher than As_{vaad}, hence enough main rebar is present.

We have mm^{2} present, so the amount of main rebar is lower than As_{vaad}, hence NOT enough main rebar is present.

Any of the following sizes and amounts will suffice. They are ranked according to surface area, from lowest to highest:

__Calculation of the shear reinforcement:__
The defining shear force is at a distace d from the edge of the support.

Hence the defining shear force is at a location d + edge of the support (which is location of the support +/- half the support width, depending on which direction you are calculating.

The defining shear force in this case (Vd) is

kN at

mm from the left support.
This is the force with which we will calculate the shear rebar spacing for the entire beam.

Another method would be to see where the minimal shear rebar is enough and tighten the gaps where the shear force grows too big.

However the method chose will provide for safer structures and will be able to take on more load capacity than a more optimized beam.

The maximum shear force capacity for our beam is calculated as follows:

V

_{Rd,max} = 0.5 * b

_{w} * d * ν * f

_{cd} =

kN.

b

_{w} = B =

mm

d =

mm

ν = 0.6 * (1-f

_{ck}/250) =

N/mm

^{2}
f

_{cd} is calculated above and is

N/mm

^{2}
The capacity of the beam to take on shear forces without any additional shear-reinforcement is calculated as follows:

V

_{Rd,c} = (C

_{Rd,c} * k * (100 * ρ

_{l} * f

_{ck})

^{1/3} * b

_{w} * d =

kN.

C

_{Rd,c} = 0.18 / γ

_{c} = 0.18 / 1.5 = 0.12

k = 1 + √(200 / d) = 1 + √(200 /

) =

ρ

_{l} = A

_{sl}/(b

_{w} * d) =

mm

^{2} / (

mm *

mm)

As V_{Rd,c} = kN > V_{d} = kN, no additional shearrebar is necessary.

The minimum required amount of shearrebar is enough. The minimum required amount is calculated as follows:

The rebar used here goes around the main rebar, so in a cross-section it appears twice.

s_{vaad} = 2*(( mm / 2) * π)^{2} / (ρ_{w,min} * b_{w}) = mm.

ρ_{w,min} = 0.08 * √(f_{ck}) / f_{yk} =

However the condition s ≤ 0.75 * d has to apply for the distance between the shear rebar.

In this case s = 0.75 * mm = mm is a more strict condition.

As a result for shear rebar we have T k.

However the condition s ≤ 0.75 * d has to apply for the distance between the shear rebar.

In this case s = 0.75 * mm = mm is a less strict condition.

As a result for shear rebar we have T k.

As V_{Rd,c} = kN < V_{d} = kN, additional shearrebar is necessary.

This is calculated by using the defining shear force Vd as a defining force for the entire beam.
s ≤ (A_{sw} * z * f_{yd} * cot(θ)) / Vd = mm

A_{sw} = 2*(( mm / 2)^{2} * π)

z = 0.9 * d

cot(θ) can be any value between 1 and 2.5. However 2.5 gives the strictest condition so that is what we use in the calculation.

As a result for shear rebar we have T k.

__Servicability State Calculation__
__Deflection__
The deflection of the beam is calculated by evaluating the effective modulus of elasticty (E) and the second moment of area (I).

The deflection of the beam is calculated based on the long-term effects of the forces acting upon it.

The second moment of area is evaluated by calculating the beam properties Xu, Xc, Iu and Ic.

For this particular problem, Xu is

mm, Iu is

mm

^{4}, Xc is

mm and Ic is

mm

^{4}.

As a result the total, positive deflection (meaning upwards) is .

And the total negative deflection is mm.

As a result the total, negative deflection (meaning downwards) is mm.

This calculated deflection is the sum of the deflection caused by the loads, shrinkage deflection and the deflection caused by creep.

With the selections made we suggest the following configuration of main-rebar:

bars of mm diameter in a single row.

You will need a single row of mm diameter,

and rows of 2 bars of mm diameter.

The diameter you selected ( mm) does not give a sensible solution.

We suggest 1 row of mm diameter.

Unfortunately for the load condition shown we have no reasonable suggestions (bars fitting in 1 row). (max diameter = 28 mm).